Harder than this one??? Wasn’t this the hardest one yet?
It seems like it is the answer, but I don’t know anyone that got it. Everyone who thought they did it “right” got 210, but that doesn’t seem right. Hoping that maybe the state sees how few people got it and, idk, give us the question back? No chance but wouldn’t that be nice.
Yeah that will be nice but the people who are grading it are not going to be nice and I heard it is getting harder day by day
Last question answer was 210
1808
Can you explain how did you get that because assuming the altitude constant and that girl being in between I tried and didn’t get the same answer
I got 321 mph too by using law of sines
It’s 210 because you had to subtract the triangle adjacent side with an angle of 15 by the adjacent side with an angle of 52 to find the distance
Everyone got 210 i know that’s right
I got 321 mph too but I used tangent
I got 210 too but now that I think about it, that’s probably wrong. It said that Freda saw the plane pass over her head, so the two triangles would be on opposite sides of Freda. I think we all missed that.
I didn’t get 210, but that def the answer. My tutor showed it to me after. When it said it’s flying over her head it doesn’t mean it’s literally over her head and she tilted her head all the way back to look at it lol. It just means it’s flying over her head, like above her. You have to do tan, because at first she saw it from a 15 degree point, and then 52. You do tan for both and subtract tan 52 from tan 15. If you drew your diagram right and carried the rest on with sin, my tutor, who checks the regents said you would get some credit.
Also the thing with the two triangles makes 0 sense. If both times she saw it was directly over her head, that means the plane didn’t even move
I just realized I did the coordinate geometry proof wrong. I used distance formula to prove congruent sides but instead of proving the diagonals perpendicular, I proved that they were congruent. The square part I’m pretty sure I got right. I proved that the slope of the sides were not perpendicular which meant the sides did not produce right angles. When I did prove the diagonals congruent, it surprisingly worked ??
Really? I’m hoping 210 is right then bc losing 3-4 points there would suck considering I knew the math.
The one thing I’m glad about is that my teacher gave us soooo many coordinate geometry proofs it was almost excessive. I proved it was a rhombus by proving it was a parallelogram with 2 congruent consecutive sides. I did the parallelogram part by showing opposite slopes congruent and then using distance formula on two adjacent sides. Then I already had the work for the slopes done to answer the square part of the question.
I used sin but still got the same answer of 18442
buddy yes it does lol
a rhombus has 4 congruent right isosceles triangles
answer was 32 pi guys
What is so difficult about the cube root problem? I was taught that in Algebra 1 not even in Geometry Honors
No it no it doesn’t have 4 isosceles because the diagonals aren’t congruent